计算此行列式 |2 1 0 0......0| |1 2 1......0 0| |0 1 2 0 0......0 0| ............ |0 0 0......12|

D 拆为两个行列式之和
1 1 0 0 ...0 1 1 0 0 ...0
0 2 1 0 ...0 1 2 1 0 ...0
0 1 2 1 ...0 + 0 1 2 1 ...0
0 0 1 2 ...0 0 0 1 2 ...0
... ... ... ...
0 0 0 0 ...2 0 0 0 0 ...2

第一个行列式按第1列展开，第二个行列式c2-c1,c3-c2,…,cn-c(n-1)
= 1 0 0 0...0
2 1 0...0 1 1 0 0...0
1 2 1...0 + 0 1 1 0...0
0 1 2...0 0 0 1 1...0
... ... ... ...
0 0 0...2 0 0 0 0...1
=D(n-1)+1
因为D1=2
所以Dn=D(n-1)+1=D(n-2)+1+1=…=D1+(n-1)=n+1